Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2010, Article ID 343580, 11 pages
doi:10.1155/2010/343580
Research Article
On Subclass of Analytic Univalent Functions
Associated with Negative Coefficients
Ma’moun Harayzeh Al-Abbadi and Maslina Darus
School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia,
Selangor D. Ehsan, Bangi 43600, Malaysia
Correspondence should be addressed to Maslina Darus, maslina@ukm.my
Received 18 November 2009; Accepted 9 January 2010
Academic Editor: Stanisława R. Kanas
Copyright q 2010 M. H. Al-Abbadi and M. Darus. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
M. H. Al-Abbadi and M. Darus 2009 recently introduced a new generalized derivative operator
, which generalized many well-known operators studied earlier by many different authors. In
μn,m
λ1 ,λ2
this present paper, we shall investigate a new subclass of analytic functions in the open unit disk
U {z ∈ C : |z| < 1} which is defined by new generalized derivative operator. Some results on
coefficient inequalities, growth and distortion theorems, closure theorems, and extreme points of
analytic functions belonging to the subclass are obtained.
1. Introduction and Definitions
Let Ax denote the class of functions of the form
fz z ∞
ak zk ,
kx1
ak is complex number,
1.1
and x ∈ N {1, 2, 3, ...}, which are analytic in the open unit disc U {z ∈ C : |z| < 1} on
the complex plane C; note that A1 A and Ax ⊆ A1. Suppose that Sx denote the
subclass of Ax consisting of functions that are univalent in U. Further, let Sα∗ x and Cα x
be the classes of Sx consisting of functions, respectively, starlike of order α and convex of
order α in U, for 0 ≤ α < 1. Let Tx denote the subclass of Sx consisting of functions of
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the form
fz z −
∞
|ak | zk ,
x ∈ N {1, 2, 3, . . .},
kx1
1.2
defined on the open unit disk U {z ∈ C : |z| < 1}. A function f ∈ Tx is called a function
with negative coefficient and the class T1 was introduced and studied by Silverman 1.
In 1 Silverman investigated the subclasses of T1 denoted by S∗T α and CT α for 0 ≤
α < 1. That are, respectively, starlike of order α and convex of order α. Now xk denotes the
Pochhammer symbol or the shifted factorial defined by
⎧
Γx k ⎨1
xk ⎩xx 1x 2 . . . x k − 1
Γx
for k 0, x ∈ C \ {0},
for k ∈ N {1, 2, . . .}, x ∈ C.
1.3
The authors in 2 have recently introduced a new generalized derivative operator
μn,m
as
follows.
λ1 ,λ2
: A → A is
Definition 1.1. For f ∈ A A1, the generalized derivative operator μn,m
λ1 ,λ2
defined by
μn,m
fz z λ1 ,λ2
∞
1 λ1 k − 1m−1
k2
1 λ2 k − 1m
where n, m ∈ N0 {0, 1, 2, . . .}, λ2 ≥ λ1 ≥ 0, and
cn, kak zk ,
cn, k nk−1 n
z ∈ U,
1.4
n 1k−1 /1k−1 .
1 Special cases of this operator include the Ruscheweyh derivative operator in the
n,0
0,m1
n
cases μn,1
≡ μn,m
≡ Sn
0,0 ≡ μ0,λ2 ≡ R 3, the Salagean derivative operator μ1,0
λ1 ,0
4, the generalized Ruscheweyh derivative operator μn,2
≡ Rnλ 5, the generalized
λ1 ,0
Salagean derivative operator introduced by Al-Oboudi μ0,m1
≡ Snβ 6, and the
λ1 ,0
n
≡ Dλ,β
where n generalized Al-Shaqsi and Darus derivative operator μλ,n1
β,0
λ, m n 1, λ1 β, and λ2 0 can be found in 7. It is easily seen that
0,0
1,1
1,1
1,m
fz μ0,m
μ0,1
0,0 fz μ0,λ2 fz μλ1 ,1 fz fz, μλ1 ,0 fz μ0,0 fz λ1 ,0
n−1,0
n−1,m
μ1,0
fz μ0,2
fz where n 1, 2, 3, . . . .
1,0 fz zf z, and also μλ1 ,0 fz μ0,0
0,λ2
By making use of the generalized derivative operator μn,m
, the authors introduce a
λ1 ,λ2
new subclass as follows.
x, α be the
Definition 1.2. For 0 ≤ α < 1, n, m ∈ N0 {0, 1, 2, . . .} and λ2 ≥ λ1 ≥ 0, let Hn,m
λ1 ,λ2
subclass of Sx consisting of functions f satisfying
⎛ ⎞
fz
z μn,m
λ1 ,λ2
⎟
⎜
Re⎝
⎠ > α,
μn,m
fz
λ1 ,λ2
z ∈ U,
1.5
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3
where
μn,m
fz z λ1 ,λ2
∞
1 λ1 k − 1m−1
cn, kak zk ,
1 λ2 k − 1m
kx1
1.6
{x 1, 2, 3, . . . , z ∈ U}, and μn,m
fz /
0.
λ1 ,λ2
n,m
Further, we define the class THλ1 ,λ2 x, α by
THn,m
α Hn,m
α ∩ Tx,
λ1 ,λ2 x,
λ1 ,λ2 x,
{x 1, 2, 3, . . .},
1.7
for 0 ≤ α < 1, n, m ∈ N0 {0, 1, 2, . . .}, and λ2 ≥ λ1 ≥ 0.
Also note that various subclasses of Hn,m
x, α and THn,m
x, α have been studied
λ1 ,λ2
λ1 ,λ2
by many authors by suitable choices of n, m, λ1 , λ2 , and x. For example,
1,1
∗
TH0,0
α ≡ TH0,1
α ≡ TH0,m
0,0 1, α ≡ THλ1 ,1 1, α ≡ ST α,
0,λ2 1,
λ1 ,0 1,
1.8
starlike of order α with negative coefficients. And
0,2
α ≡ TH1,1
α ≡ TH1,m
TH1,0
0,0 1, α ≡ TH1,0 1, α ≡ CT α,
0,λ2 1,
λ1 ,0 1,
1.9
class of convex function of order α with negative coefficients. Also
1,1
∗
α ≡ TH0,1
α ≡ TH0,m
TH0,0
0,0 x, α ≡ THλ1 ,1 x, α ≡ ST x, α,
0,λ2 x,
λ1 ,0 x,
0,2
TH1,0
α ≡ TH1,1
α ≡ TH1,m
0,0 x, α ≡ TH1,0 x, α ≡ CT x, α,
0,λ2 x,
λ1 ,0 x,
TH0,2
x, α ≡ P x, λ1 , α
λ1 ,0
0 ≤ λ1 < 1,
TH1,2
α ≡ Cx, λ1 , α
λ1 ,0 x,
0 ≤ λ1 < 1.
1.10
The classes S∗T x, α and CT x, α were studied by Chatterjea 8 see also Srivastava et al.
9, whereas the classes P x, λ1 , α and Cx, λ1 , α were, respectively, studied by Altintaş
10 and Kamali and Akbulut 11. When λ1 λ2 0 or m 1, λ2 0, or m 0, λ1 0 in the
x, α, we have the class Rn α introduced and studied by Ahuja 12. Finally we
class Hn,m
λ1 ,λ2
note that when m 2, λ2 0 in the class Hn,m
x, α we have the class Knλ x, α introduced
λ1 ,λ2
and studied by Al-Shaqsi and Darus 13.
2. Coefficient Inequalities
In this section, we provide a necessary and sufficient condition for a function f analytic in U
x, α and in THn,m
x, α.
to be in Hn,m
λ1 ,λ2
λ1 ,λ2
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International Journal of Mathematics and Mathematical Sciences
Theorem 2.1. For 0 ≤ α < 1 and λ2 ≥ λ1 ≥ 0, let f ∈ Sx be defined by 1.1. If
∞
k − α1 λ1 k − 1m−1
cn, k|ak | ≤ 1 − α,
1 λ2 k − 1m
kx1
x 1, 2, . . . ,
2.1
then f ∈ Hn,m
x, α, where n ∈ N {1, 2, . . .} and m ∈ N0 {0, 1, 2, . . .}.
λ1 ,λ2
Proof. Assume that 2.1 holds true. Then we shall prove condition 1.5. It is sufficient to
show that
z μn,m
fz
λ1 ,λ2
−
1
≤ 1 − α,
μn,m fz
λ1 ,λ2
2.2
z ∈ U.
So, we have that
n,m
n,m
z μn,m
z
μ
fz
fz
−
μ
fz
λ1 ,λ2
λ1 ,λ2
λ1 ,λ2
− 1 n,m
n,m
μ
fz
μ
fz
λ1 ,λ2
λ1 ,λ2
∞ kx1 k − 11 λ1 k − 1m−1 /1 λ2 k − 1m cn, kak zk m−1
m
z ∞
/1 λ2 k − 1 cn, kak zk kx1 1 λ1 k − 1
|z| < 1,
∞
≤
k − 11 λ1 k − 1m−1 /1 λ2 k − 1m cn, k|ak |
,
m−1
1− ∞
/1 λ2 k − 1m cn, k|ak |
kx1 1 λ1 k − 1
kx1
2.3
and expression 2.3 is bounded by 1 − α.
Hence 2.2 holds if
∞
k − 11 λ1 k − 1m−1
cn, k|ak |
1 λ2 k − 1m
kx1
∞
1 λ1 k − 1m−1
≤ 1 − α 1 −
cn, k|ak | ,
1 λ2 k − 1m
kx1
2.4
which is equivalent to
∞
k − α1 λ1 k − 1m−1
cn, k|ak | ≤ 1 − α,
1 λ2 k − 1m
kx1
2.5
International Journal of Mathematics and Mathematical Sciences
5
x, α. Note that the denominator in 2.3 is positive provided that
by 2.1. Thus f ∈ Hn,m
λ1 ,λ2
2.1 holds.
m−1
Theorem 2.2. Let f be defined by 1.2 and 1 − ∞
/1 λ2 k − 1m c(n,
kx1 k1 λ1 k − 1
n,m
k)|ak | ≥ 0 x 1, 2, . . .. Then f ∈ THλ1 ,λ2 x, α if and only if 2.1 is satisfied.
Proof. We only prove the right-hand side, since the other side can be justified using similar
x, α by condition 1.5, we have that
arguments in proof of Theorem 2.1. Since f ∈ THn,m
λ1 ,λ2
⎛ ⎞
fz
z μn,m
λ1 ,λ2
⎟
⎜
Re⎝
⎠
n,m
μλ1 ,λ2 fz
⎧
⎫
m−1
⎨z − ∞
/1 λ2 k − 1m cn, k|ak |zk ⎬
kx1 k1 λ1 k − 1
> α.
Re
m−1
⎩ z − ∞
/1 λ2 k − 1m cn, k|ak |zk ⎭
kx1 1 λ1 k − 1
2.6
Choose values of z on real axis so that zμn,m
fz /μn,m
fz is real. Letting z → 1− through
λ1 ,λ2
λ1 ,λ2
real values, we have that
k1 λ1 k − 1m−1 /1 λ2 k − 1m cn, k|ak |zk
> α.
m−1
m
k
cn,
k|a
1− ∞
/1
λ
−
1
−
1
λ
k
k
1
|z
1
2
k
kx1
1−
∞
kx1
2.7
Thus we obtain
∞
k − α1 λ1 k − 1m−1
cn, k|ak | ≤ 1 − α,
1 λ2 k − 1m
kx1
2.8
which is 2.1. Hence the proof is complete.
The result is sharp with the extremal function f given by
fz z −
1 − α1 λ2 xm
x 1 − α1 λ1 xm−1 cn, x 1
zx1 ,
x 1, 2, . . ..
2.9
x, α. Then
Theorem 2.3. Let the function f given by 1.2 be in the class THn,m
λ1 ,λ2
|ak | ≤
1 − α1 λ2 k − 1m
k − α1 λ1 k − 1m−1 cn, k
,
2.10
where 0 ≤ α < 1, λ2 ≥ λ1 ≥ 0, k ≥ x 1, and x 1, 2, 3, . . . . Equality holds for the function given
by 2.9.
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International Journal of Mathematics and Mathematical Sciences
x, α, then condition 2.1 gives
Proof. Since f ∈ THn,m
λ1 ,λ2
|ak | ≤
1 − α1 λ2 k − 1m
k − α1 λ1 k − 1m−1 cn, k
,
2.11
for each k x 1 where x 1, 2, 3, . . . .
Clearly the function given by 2.9 satisfies 2.10, and therefore, f given by 2.9 is in
n,m
THλ1 ,λ2 x, α for this function; the result is clearly sharp.
3. Growth and Distortion Theorems
In this section, growth and distortion theorems will be considered and covering property for
function in the class will also be given.
x, α. Then for 0 < |z| r <
Theorem 3.1. Let the function f given by 1.2 be in the class THn,m
λ1 ,λ2
1,
r−
1 − α1 λ2 xm
x 1 − α1 λ1 xm−1 cn, x 1
≤ fz ≤ r r x1
3.1
1 − α1 λ2 xm
x 1 − α1 λ1 xm−1 cn, x 1
r
x1
,
x 1, 2, . . . ,
where 0 ≤ α < 1, m ∈ N0 {0, 1, 2 . . .}, and n ∈ N {1, 2, . . .}.
Proof. We only prove the right-hand side inequality in 3.1, since the other inequality can be
x, α by Theorem 2.2, we have that
justified using similar arguments. Since f ∈ THn,m
λ1 ,λ2
∞
k − α1 λ1 k − 1m−1
cn, k|ak | ≤ 1 − α.
1 λ2 k − 1m
kx1
3.2
Now
x 1 − α1 λ1 xm−1
cn, x 1
1 λ2 xm
∞
|ak |
kx1
∞
x 1 − α1 λ1 xm−1
cn, x 1|ak |,
1 λ2 xm
kx1
≤
∞
k − α1 λ1 k − 1m−1
cn, k|ak |,
1 λ2 k − 1m
kx1
≤ 1 − α,
x 1, 2, 3, . . . .
3.3
International Journal of Mathematics and Mathematical Sciences
7
And therefore,
∞
|ak | ≤
kx1
1 − α1 λ2 xm
x 1 − α1 λ1 xm−1 cn, x 1
,
x 1, 2, . . . .
3.4
Since
fz z −
∞
|ak |zk ,
x 1, 2, . . . ,
3.5
kx1
then we have that
∞
fz z −
|ak |zk .
kx1
3.6
After that,
∞
fz ≤ |z| |z|x1
|ak ||z|k−x1
kx1
≤rr
x1
∞
3.7
|ak |.
kx1
By aid of inequality 3.4, it yields the right-hand side inequality of 3.1. Thus, this completes
the proof.
Theorem 3.2. Let the function f given by 1.2 be in the class THn,m
x, α. Then for 0 < |z| r <
λ1 ,λ2
1,
1−
x 11 − α1 λ2 xm
x 1 − α1 λ1 xm−1 cn, x 1
≤ f z ≤ 1 rx
x 11 − α1 λ2 xm
x 1 − α1 λ1 xm−1 cn, x 1
3.8
x
r ,
x 1, 2, . . . ,
where 0 ≤ α < 1, λ2 ≥ λ1 ≥ 0, m ∈ N0 {0, 1, 2, . . .}, and n ∈ N {1, 2, . . .}.
Proof. Since f ∈ THn,m
x, α, by Theorem 2.2, we have that
λ1 ,λ2
∞
k − α1 λ1 k − 1m−1
cn, k|ak | ≤ 1 − α,
1 λ2 k − 1m
kx1
x 1, 2, 3, . . . .
3.9
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International Journal of Mathematics and Mathematical Sciences
Now
x 1 − α1 λ1 xm−1
cn, x 1
1 λ2 xm
∞
k|ak |
kx1
∞
x 1 − α1 λ1 xm−1
cn, x 1k|ak |
1 λ2 xm
kx1
≤ x 1
3.10
∞
k − α1 λ1 k − 1m−1
cn, k|ak |
1 λ2 k − 1m
kx1
≤ x 11 − α,
k ≥ x 1, x 1, 2, 3, . . ..
Hence
∞
x 11 − α1 λ2 xm
k|ak | ≤
x 1 − α1 λ1 xm−1 cn, x 1
kx1
,
x 1, 2, . . . .
3.11
Since
f z 1 −
∞
k|ak |zk−1 ,
x 1, 2, . . . ,
3.12
kx1
then we have that
1 − |z|x
∞
∞
k|ak ||z|k−x1 ≤ f z ≤ 1 |z|x
k|ak ||z|k−x1 ,
kx1
kx1
3.13
and therefore,
1 − rx
∞
∞
k|ak | ≤ f z ≤ 1 r x
k|ak |,
kx1
x 1, 2, . . . .
kx1
3.14
By using the inequality 3.11 in 3.14, we get Theorem 3.2. This completes the proof.
4. Extreme Points
The extreme points of the class THn,m
x, α are given by the following theorem.
λ1 ,λ2
Theorem 4.1. Let fx z z and
fk z z −
1 − α1 λ2 k − 1m
k − α1 λ1 k − 1m−1 cn, k
zk ,
where 0 ≤ α < 1, λ2 ≥ λ1 ≥ 0, n, m ∈ N0 , k x 1, x 2, . . . , and x 1, 2, 3, . . . .
4.1
International Journal of Mathematics and Mathematical Sciences
9
x, α if and only if it can be expressed in the form
Then f ∈ THn,m
λ1 ,λ2
fz ∞
δk fk z,
4.2
kx
where δk ≥ 0 and
∞
kx
δk 1.
Proof. Suppose that f can be expressed as in 4.2. Our goal is to show that f ∈ THn,m
x, α.
λ1 ,λ2
By 4.2, we have that
fz ∞
δk fk z
kx
∞
δx fx z δk fk z
kx1
∞
δx fx z δk z −
kx1
∞
δk z −
kx
z−
∞
k − α1 λ1 k − 1m−1 cn, k
δk 1 − α1 λ2 k − 1m
kx1 k
∞
1 − α1 λ2 k − 1m
− α1 λ1 k − 1m−1 cn, k
δk 1 − α1 λ2 k − 1m
kx1 k
− α1 λ1 k − 1m−1 cn, k
k
z
4.3
zk
zk .
Now
fz z −
∞
|ak |zk z −
kx1
∞
δk 1 − α1 λ2 k − 1m
m−1
cn, k
kx1 k − α1 λ1 k − 1
zk ,
4.4
so that
δk 1 − α1 λ2 k − 1m
|ak | k − α1 λ1 k − 1m−1 cn, k
.
4.5
Now, we have that
∞
δk 1 − δx ≤ 1,
x 1, 2, 3, . . . .
kx1
4.6
Setting
∞
δk kx1
∞
δk 1 − α1 λ2 k − 1m
k − α1 λ1 k − 1m−1 cn, k
≤ 1,
m−1
1 − α1 λ2 k − 1m
cn, k
kx1 k − α1 λ1 k − 1
4.7
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International Journal of Mathematics and Mathematical Sciences
we arrive to
∞
k − α1 λ1 k − 1m−1
cn, k|ak | ≤ 1.
1 − α1 λ2 k − 1m
kx1
4.8
And therefore,
∞
k − α1 λ1 k − 1m−1
cn, k|ak | ≤ 1 − α,
1 λ2 k − 1m
kx1
x 1, 2, 3, . . . .
4.9
It follows from Theorem 2.2 that f ∈ THn,m
x, α.
λ1 ,λ2
x, α; our goal is, to get 4.2. From 4.2
Conversely, let us suppose that f ∈ THn,m
λ1 ,λ2
and using similar last arguments, it is easily seen that
fz z −
∞
|ak |zk z −
kx1
∞
kx1 k
δk 1 − α1 λ2 k − 1m
− α1 λ1 k − 1m−1 cn, k
zk ,
4.10
which suffices to show that
|ak | δk 1 − α1 λ2 k − 1m
k − α1 λ1 k − 1m−1 cn, k
.
4.11
Now, we have that f ∈ THn,m
x, α, then by previous Theorem 2.3,
λ1 ,λ2
1 − α1 λ2 k − 1m
.
4.12
k − α1 λ1 k − 1m−1 cn, k|ak |
≤ 1.
1 − α1 λ2 k − 1m
4.13
|ak | ≤
k − α1 λ1 k − 1m−1 cn, k
That is
Since
∞
kx δk 1, we see δk ≤ 1, for each k x, x 1, x 2, . . . , and x 1, 2, 3, . . . .
We can set that
δk k − α1 λ1 k − 1m−1 cn, k|ak |
.
1 − α1 λ2 k − 1m
4.14
Thus, the desired result is that
|ak | δk 1 − α1 λ2 k − 1m
k − α1 λ1 k − 1m−1 cn, k
This completes the proof of the theorem.
.
4.15
International Journal of Mathematics and Mathematical Sciences
11
x, α are the functions
Corollary 4.2. The extreme points of THn,m
λ1 ,λ2
fx z z,
fk z z −
1 − α1 λ2 k − 1m
k − α1 λ1 k − 1
m−1
cn, k
zk ,
4.16
where 0 ≤ α < 1, n, m ∈ N0 {0, 1, 2, . . .}, λ2 ≥ λ1 ≥ 0, and k x1, x2, . . . , x 1, 2, 3, . . ..
Acknowledgments
This work is fully supported by UKM-GUP-TMK-07-02-107, Malaysia. The authors are also
grateful to the referee for his/her suggestions which helped us to improve the contents of this
article.
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